Warning: mysql_fetch_array():

phence

this code gives me this error!?!
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/www/chrisdavis/Projects/search.php on line 37
Why? Thanks!

<?		
$connection = @ mysql_pconnect($hostName, $username, $password) or die(mysql_error());
		$db = mysql_select_db($databaseName, $connection) or die(mysql_error());

	$query = mysql_query("SELECT (pro_image1, pro_name) FROM projects WHERE (pro_feat) = ('yes')");
/* Line 37 */	while ($feature = mysql_fetch_array($query)){ 
  		?>
	  	<td><? $feature['pro_image1']; ?></td></tr>
	  	<tr><td><? $feature['pro_name']; ?></td>
		<?
	} // End While
?>

rachel2004

Use mysql_error() for troubleshooting. Try this:

$query = mysql_query("SELECT pro_image1, pro_name FROM projects WHERE pro_feat = 'yes'", $connection)
    or die("<p>Couldn't execute query: <strong>".mysql_error()."</strong></p>");

pablodelapena

maybe its just me. but this line looks odd:

$query = mysql_query("SELECT (pro_image1, pro_name) FROM projects WHERE (pro_feat) = ('yes')");

Is it normal practice to have brackets?! Please exscuse me if im being an idiot.

phence

yeah.. it's my bad.. havn't wrote a sql line in a couple of months, forgot about not using it when i've got quotes and no variable